Answer
$0,\frac{\pi}{6},\frac{5\pi}{6}$
Work Step by Step
1. $sin(2\theta)-cos(\theta)-2sin(\theta)+1=0 \Longrightarrow 2sin(\theta)cos(\theta)-cos(\theta)-2sin(\theta)+1=0 \Longrightarrow cos(\theta)(2sin(\theta)-1)-(2sin(\theta)-1)=0 \Longrightarrow (2sin(\theta)-1)(cos(\theta)-1)=0 \Longrightarrow sin(\theta)=\frac{1}{2},\ or\ cos(\theta)=1$,
2. For $cos(\theta)=1$, we have $\theta=2k\pi$
3. For $sin(\theta)=\frac{1}{2}$, we have $\theta=2k\pi+\frac{\pi}{6}$ or $\theta=2k\pi+\frac{5\pi}{6}$
4. Within $[0,2\pi)$, we have $\theta=0,\frac{\pi}{6},\frac{5\pi}{6}$